Projectile Motion Lab Report
Physics 223-101
Atwood’s Machine
DATA TABLE Part 1: Keeping Total Mass Constant | Trial | m1(g) | m2(g) | Acceleration(m/s2) | Δm(kg) | mT(kg) | 1 | 200 | 200 | 0 | 0 | 0.400 | 2 | 205 | 195 | -0.174 | 0.01 | 0.400 | 3 | 210 | 190 | -0.382 | 0.02 | 0.400 | 4 | 215 | 185 | -0.607 | 0.03 | 0.400 | 5 | 220 | 180 | -0.830 | 0.04 | 0.400 | | | | | | | Part II: Keeping the Mass Difference Constant | Trial | m1(g) | m2(g) | Acceleration(m/s2) | Δm(kg) | mT(kg) | 1 | 120 | 100 | -0.610 | 0.02 | 0.220 | 2 | 140 | 120 | -0.590 | 0.02 | 0.260 | 3 | 160 | 140 | -0.520 | 0.02 | 0.300 | 4 | 180 | 160 | -0.459 | 0.02 | 0.340 | 5 | 200 | 180 | -0.414 | 0.02 | 0.380 |
Calculations: H = 0.97 m θ = 0° a = 9.81 m/s2 Time: t=2ya t= 2(0.97m)9.81m/s2 t= 0.44 s Davg= (D1+D2+D3+D4+D5)/n = (2.146m+2.235m+2.234m+2.222m+2.225m)/5 = 2.212 m
Vox= Davgt Vy = Vosinθ-gt Vo = gxsinθcosθ = 2.214m0.44s = 0-(9.81m/s2)(0.44s) = 9.81(2.62)sin45(cos45) = 5.032 m/s = 4.312 m/ = 7.17 m/s Discussion:
The two pictures below demonstrate the calculations from Trial 1 at 0° and Trial 1 at 45°. In this experiment we did five separate shots from the cannon at 0°, and 3 separate shots at 45°. We determined the average distance of the shots from each angle by using a meter stick. We were then able to calculate the velocity of the shots by dividing the time from distance. After getting the results, the distance from the shots fired at 45° had a greater distance from the shots fired at 0° horizontally. This is because as the ball which is fired increases, the height increases. As the height increases, the time it takes for the ball to hit the ground increases and the range (distance) is greater.
Conclusion:
One source of error comes from when we shot the cannon from the ground at a
Atwood’s Machine
DATA TABLE Part 1: Keeping Total Mass Constant | Trial | m1(g) | m2(g) | Acceleration(m/s2) | Δm(kg) | mT(kg) | 1 | 200 | 200 | 0 | 0 | 0.400 | 2 | 205 | 195 | -0.174 | 0.01 | 0.400 | 3 | 210 | 190 | -0.382 | 0.02 | 0.400 | 4 | 215 | 185 | -0.607 | 0.03 | 0.400 | 5 | 220 | 180 | -0.830 | 0.04 | 0.400 | | | | | | | Part II: Keeping the Mass Difference Constant | Trial | m1(g) | m2(g) | Acceleration(m/s2) | Δm(kg) | mT(kg) | 1 | 120 | 100 | -0.610 | 0.02 | 0.220 | 2 | 140 | 120 | -0.590 | 0.02 | 0.260 | 3 | 160 | 140 | -0.520 | 0.02 | 0.300 | 4 | 180 | 160 | -0.459 | 0.02 | 0.340 | 5 | 200 | 180 | -0.414 | 0.02 | 0.380 |
Calculations: H = 0.97 m θ = 0° a = 9.81 m/s2 Time: t=2ya t= 2(0.97m)9.81m/s2 t= 0.44 s Davg= (D1+D2+D3+D4+D5)/n = (2.146m+2.235m+2.234m+2.222m+2.225m)/5 = 2.212 m
Vox= Davgt Vy = Vosinθ-gt Vo = gxsinθcosθ = 2.214m0.44s = 0-(9.81m/s2)(0.44s) = 9.81(2.62)sin45(cos45) = 5.032 m/s = 4.312 m/ = 7.17 m/s Discussion:
The two pictures below demonstrate the calculations from Trial 1 at 0° and Trial 1 at 45°. In this experiment we did five separate shots from the cannon at 0°, and 3 separate shots at 45°. We determined the average distance of the shots from each angle by using a meter stick. We were then able to calculate the velocity of the shots by dividing the time from distance. After getting the results, the distance from the shots fired at 45° had a greater distance from the shots fired at 0° horizontally. This is because as the ball which is fired increases, the height increases. As the height increases, the time it takes for the ball to hit the ground increases and the range (distance) is greater.
Conclusion:
One source of error comes from when we shot the cannon from the ground at a